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Thread: Question involving a collection of closed intervals

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    Question involving a collection of closed intervals

    Let $\displaystyle F$ be the collection of closed intervals $\displaystyle A_n=[1/n, 1-1/n]$ for $\displaystyle n=3,4,5,...$. What do you notice about $\displaystyle \bigcup F$? Is it closed, open, both, or neither?

    I'm afraid the wording and available material has got me stumped on this, and being sick at the moment doesn't help.
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    Notice that $\displaystyle \left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$.
    Can either 1 or 0 be included?
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    Quote Originally Posted by Plato View Post
    Notice that $\displaystyle \left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$.
    Can either 1 or 0 be included?
    I should note, the original question said that $\displaystyle n=1,2,3,...$, but our Prof. told us to change it to $\displaystyle n=3,4,5,...$ when he realized that when $\displaystyle n=1, A_n=[1,0]$ and when $\displaystyle n=2, A_n=[1/2,1/2]$ (or at least that's what I heard him say). I wrote the question word-for-word (besides the change our Prof. gave us), so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.
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    Quote Originally Posted by Runty View Post
    so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.
    That is the whole point: neither 0 nor 1 can be included.
    So what is the union?
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    Quote Originally Posted by Plato View Post
    That is the whole point: neither 0 nor 1 can be included.
    So what is the union?
    I am presently guessing, but here is my answer thus far.

    $\displaystyle \bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$

    My guess so far indicates that the union is both open and closed, though I could easily be wrong.
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    Quote Originally Posted by Runty View Post
    I am presently guessing, but here is my answer thus far.

    $\displaystyle \bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$
    For any $\displaystyle k$ does $\displaystyle 1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
    If not how can the union be $\displaystyle (0,1]?$
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    Quote Originally Posted by Plato View Post
    For any $\displaystyle k$ does $\displaystyle 1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
    If not how can the union be $\displaystyle (0,1]?$
    Point taken. I guess it's a completely open interval, then.

    EDIT: One more thing, I've looked at some work done on this in another forum, for which I will provide the link.

    http://www.mymathforum.com/viewtopic...=15805&start=0

    Is the work found here accurate?
    Last edited by Runty; Sep 19th 2010 at 12:31 PM.
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